Question

A student performs an experiment to determine the Young’s modulus of a wire, exactly 2 m long, by Searle’s method. In a particular reading, the students measures the extension in the length of the wire to be 0.8 mm with an uncertainty of +/- 0.05 mm at a load of exactly 1kg. The student also measures the diameter of the wire to be 0.4 mm with an uncertainty of +/-0.01 mm.The Young’s modulus obtained from the reading is $(g=9.8m{s}^{-2})$

• A. (2.0 +/- 0.3) X 10¹¹ Nm^–2
• B. (2.0 +/- 0.2) X 10¹¹ Nm^–2
• C. (2.0 +/- 0.1) X 10¹¹ Nm^–2
• D. (2.0 +/- 0.05) X 10¹¹ Nm^–2

Answer 1

The correct option is B (2.0 +/- 0.2) X 10¹¹ Nm^–2

$Y=\frac{4Fl}{\pi D^2△l}$
$=\frac{4\times \left(1X9.8\right)X2}{\left(\frac{22}{7}\right)(0.4\times {10}^{-}3{)}^{2}X(0.8X{10}^{-3})}$
$=2\times {10}^{11}N{m}^{-2}$
Fractional error in y is
$\frac{△y}{y}=+/-(\frac{2△D}{D}+\frac{△l}{l})$
$or△y=+/-(\frac{2X0.01}{0.4}+\frac{0.05}{0.8})\times 2\times {10}^{11}$
$=+/-0.2\times {10}^{11}N{m}^{-2}$
Therefore $y=(2\times {10}^{11}+/-0.2\times {10}^{11})N{m}^{-2}$